一开始看到题还以为要DFS还是BFS，后来发现完全不用。排个序，然后看看大小王能不能弥补缺口就行，但后来又发现还要排除有相同大小牌的情况。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std; int main() {
int n;
while (cin >> n && n != 0) {
vector<int> cards;
int kingCount = 0;
while (n--) {
int x;
cin >> x;
if (x == 0) {
kingCount++;
}
else {
cards.push_back(x);
}
}
sort(cards.begin(), cards.end());
bool lucky = true;
for (int i = 1; i < cards.size(); i++) {
if (cards[i] == cards[i-1]) {
lucky = false;
}
kingCount -= cards[i] - cards[i-1] - 1;
if (kingCount < 0) {
lucky = false;
}
}
if (lucky) {
cout << "So Lucky!" << endl;
}
else {
cout << "Oh My God!" << endl;
}
}
}